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At time t = 0, a projectile was fired up...

At time `t = 0`, a projectile was fired upward from an initial height of 10 feet. Its height after t seconds is given by the function `h(t) = p - 10(q - t)^2`, where p and q are positive constants. If the projectile reached a maximum height of 100 feet when t = 3, then what was the height, in feet, of the projectile when t = 4 ?

A

62

B

70

C

85

D

90

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we begin with the given height function of the projectile: \[ h(t) = p - 10(q - t)^2 \] ### Step 1: Use the initial condition At time \( t = 0 \), the height of the projectile is given as 10 feet. We can substitute \( t = 0 \) into the height function: \[ h(0) = p - 10(q - 0)^2 = 10 \] This simplifies to: \[ p - 10q^2 = 10 \] This is our **Equation 1**. ### Step 2: Use the maximum height condition We know that the projectile reaches a maximum height of 100 feet at \( t = 3 \). We substitute \( t = 3 \) into the height function: \[ h(3) = p - 10(q - 3)^2 = 100 \] Expanding this, we have: \[ h(3) = p - 10(q^2 - 6q + 9) = 100 \] This simplifies to: \[ p - 10q^2 + 60q - 90 = 100 \] Rearranging gives us: \[ p - 10q^2 + 60q - 190 = 0 \] This is our **Equation 2**. ### Step 3: Solve the equations Now we have two equations: 1. \( p - 10q^2 = 10 \) (Equation 1) 2. \( p - 10q^2 + 60q - 190 = 0 \) (Equation 2) From Equation 1, we can express \( p \) in terms of \( q \): \[ p = 10 + 10q^2 \] Substituting this into Equation 2: \[ (10 + 10q^2) - 10q^2 + 60q - 190 = 0 \] This simplifies to: \[ 10 + 60q - 190 = 0 \] So, \[ 60q = 180 \] Dividing both sides by 60 gives: \[ q = 3 \] ### Step 4: Find \( p \) Now substituting \( q = 3 \) back into Equation 1 to find \( p \): \[ p - 10(3^2) = 10 \] This simplifies to: \[ p - 90 = 10 \] Thus, \[ p = 100 \] ### Step 5: Calculate the height at \( t = 4 \) Now we have \( p = 100 \) and \( q = 3 \). We need to find the height at \( t = 4 \): \[ h(4) = p - 10(q - 4)^2 \] Substituting the values of \( p \) and \( q \): \[ h(4) = 100 - 10(3 - 4)^2 \] Calculating \( (3 - 4)^2 \): \[ (3 - 4)^2 = (-1)^2 = 1 \] So, \[ h(4) = 100 - 10(1) = 100 - 10 = 90 \] ### Final Answer The height of the projectile when \( t = 4 \) seconds is: \[ \boxed{90} \text{ feet} \]
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